If the system of equations  \begin{align*}
3x+y&=a,\\
2x+5y&=2a,
\end{align*} has a solution $(x,y)$ when $x=2$, compute $a$.
Substituting in $x=2$, we obtain the equations

\begin{align*}
y+6&=a,\\
5y+4&=2a.
\end{align*}

Multiplying the first equation by $5$ and subtracting it from the second equation, we find

$$-26=-3a\Rightarrow a=\boxed{\frac{26}{3}}.$$